Turn on thread page Beta
    • Thread Starter
    Offline

    11
    ReputationRep:
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    Offline

    9
    ReputationRep:
    Look this up on Wolfram alpha
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by emperorCode)
    Look this up on Wolfram alpha
    I tried but they don't differentiate it into the way I want.
    Offline

    17
    ReputationRep:
    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    Separate into 0.5^0.5 * (x^3-4x+4)^1/2, as 0.5^0.5 is a constant it shouldnt hinder ur ability to differentiate the other expression. Just remember to multiply by 0.5^0.5 after uve differentiated (x^3-4x+4)^1/2.
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Shaanv)
    Separate into 0.5^0.5 * (x^3-4x+4)^1/2, as 0.5^0.5 is a constant it shouldnt hinder ur ability to differentiate the other expression. Just remember to multiply by 0.5^0.5 after uve differentiated (x^3-4x+4)^1/2.
    So would this give me my differential function?
    Offline

    17
    ReputationRep:
    (Original post by mansnothot)
    So would this give me my differential function?
    This will give u the gradient function. Are u talking about a specific question or just in general in ur intial post?
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by Shaanv)
    This will give u the gradient function. Are u talking about a specific question or just in general in ur intial post?
    no i just needed to differentiate that because I didnt know how to. If it's okay can I attempt to differentiate it and could you tell me if I'm correct?

    Thanks.
    Offline

    6
    ReputationRep:
    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
    Offline

    17
    ReputationRep:
    (Original post by RickiestRick)
    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
    SMH, change ur name to mortiest morty. 😂
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    6
    ReputationRep:
    (Original post by Shaanv)
    SMH, change ur name to mortiest morty. 😂
    But I'm arrogant, and that's exactly who Rick is
    Offline

    19
    ReputationRep:
    (Original post by RickiestRick)
    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
    That's such a dank unit!

    Wait you do new spec

    e . w
    Offline

    9
    ReputationRep:
    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    nice username g
    Offline

    15
    ReputationRep:
    (Original post by perspirationting)
    nice username g
    Loooooooooool I'm done
    Offline

    6
    ReputationRep:
    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    You use the chain rule for the first question and the quotient rule for the second.

    Not sure if this is allowed but I've attached an explanation of how to use the chain rule regarding the first question. Hope it makes sense and helps.

    Spoiler:
    Show


    Name:  IMG_20171026_224606692.jpg
Views: 36
Size:  493.0 KB

    For some bizarre reason, I can't upload the final answer, which is 1/3(6x + 5)(3x^2 + 5 - 9)^-2/3)

    Attached Images
      
    Offline

    7
    ReputationRep:
    Is it this...
    Spoiler:
    Show

    y = (½(x^3 - 4x + 4))^½
    = ½^½ * (x^3 - 4x + 4)^½
    dy/dx = ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    = ½^3/2 * (x^3 - 4x + 4)^-½ * (3x^2 - 4)

    ∴ = ½^3/2 (x^3 - 4x + 4)^-½(3x^2 - 4)

    • Thread Starter
    Offline

    11
    ReputationRep:
    (Original post by amkay)
    Is it this...
    Spoiler:
    Show



    y = (½(x^3 - 4x + 4))^½
    = ½^½ * (x^3 - 4x + 4)^½
    dy/dx = ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    = ½^3/2 * (x^3 - 4x + 4)^-½ * (3x^2 - 4)

    ∴ = ½^3/2 (x^3 - 4x + 4)^-½(3x^2 - 4)



    I got up to this part ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    but I don't understand why ½^½ * ½ becomes ½^3/2
    Offline

    7
    ReputationRep:
    (Original post by mansnothot)
    I got up to this part ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    but I don't understand why ½^½ * ½ becomes ½^3/2
    I'm just adding powers so ½^½ * ½^1 = ½^(½ + 1) = ½^3/2
    just like how x^2 * x^3 = x^(2+3) = x^5
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 28, 2017

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.